calculus rogers.pdf

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The Calculus of Several Variables

Robert C. Rogers

September 29, 2011


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It is now known to science that there are many more dimensions than

the classical four. Scientists say that these dont normally impinge on

the world because the extra dimensions are very small and curve in onthemselves, and that since reality is fractal most of it is tucked inside

itself. This means either that the universe is more full of wonders than

we can hope to understand or, more probably, that scientists make

things up as they go along.

Terry Pratchett

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Contents

1 Introduction 1

I Precalculus of Several Variables 5

2 Vectors, Points, Norm, and Dot Product 6

3 Angles and Projections 14

4 Matrix Algebra 19

5 Systems of Linear Equations and Gaussian Elimination 27

6 Determinants 38

7 The Cross Product and Triple Product inR3 47

8 Lines and Planes 55

9 Functions, Limits, and Continuity 60

10 Functions from R to Rn 70

11 Functions from Rn to R 76

12 Functions from Rn to Rm 81

12.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8112.2 Parameterized Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 84

13 Curvilinear Coordinates 86

13.1 Polar Coordinates inR2

. . . . . . . . . . . . . . . . . . . . . . . 8613.2 Cylindrical Coordinates in R3 . . . . . . . . . . . . . . . . . . . . 8813.3 Spherical Coordinates inR3 . . . . . . . . . . . . . . . . . . . . . 90

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II Differential Calculus of Several Variables 93

14 Introduction to Differential Calculus 94

15 Derivatives of Functions from R to Rn 96

16 Derivatives of Functions from Rn to R 101

16.1 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 10116.2 Higher Order Partial Derivatives . . . . . . . . . . . . . . . . . . 10216.3 The Chain Rule for Partial Derivatives . . . . . . . . . . . . . . . 105

17 Derivatives of Functions from Rn to Rm 113

17.1 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 11317.2 The Total Derivative Matrix . . . . . . . . . . . . . . . . . . . . . 11417.3 The Chain Rule for Mappings . . . . . . . . . . . . . . . . . . . . 118

18 Gradient, Divergence, and Curl 123

18. 1 The Gradi ent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12318. 2 The Di vergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12718.3 T he Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

19 Differential Operators in Curvilinear Coordinates 133

19.1 Differential Operators Polar Coordinates . . . . . . . . . . . . . . 13319.2 Differential Operators in Cylindrical Coordinates . . . . . . . . . 13719.3 Differential Operators in Spherical Coordinates . . . . . . . . . . 139

20 Differentiation Rules 142

20.1 Linearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14220. 2 Product Rul es . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14220.3 Second Derivative Rules . . . . . . . . . . . . . . . . . . . . . . . 143

21 Eigenvalues 146

22 Quadratic Approximation and Taylors Theorem 157

22.1 Quadratic Approximation of Real-Valued Functions . . . . . . . 15722.2 Taylors Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

23 Max-Min Problems 165

23.1 First Derivative Test . . . . . . . . . . . . . . . . . . . . . . . . . 16823.2 Second Derivative Test . . . . . . . . . . . . . . . . . . . . . . . . 17023.3 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . 175

24 Nonlinear Systems of Equations 18124.1 The Inverse Function Theorem . . . . . . . . . . . . . . . . . . . 18124.2 The Implicit Function Theorem . . . . . . . . . . . . . . . . . . . 184

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I II Integral Calculus of Several Variables 190

25 Introduction to Integral Calculus 191

26 Riemann Volume in Rn 195

27 Integrals Over Volumes in Rn 199

27.1 Basic Definitions and Properties . . . . . . . . . . . . . . . . . . 19927.2 Basic Properties of the Integral . . . . . . . . . . . . . . . . . . . 20127.3 Integrals Over Rectangular Regions . . . . . . . . . . . . . . . . . 20327.4 Integrals Over General Regions in R2 . . . . . . . . . . . . . . . . 20527.5 Change of Order of Integration in R2 . . . . . . . . . . . . . . . . 20827.6 Integrals over Regions inR3 . . . . . . . . . . . . . . . . . . . . . 211

28 The Change of Variables Formula 217

29 Hausdorff Dimension and Measure 231

30 Integrals over Curves 235

30.1 The Length of a Curve . . . . . . . . . . . . . . . . . . . . . . . . 23530.2 Integrals of Scalar Fields Along Curves . . . . . . . . . . . . . . . 23730.3 Integrals of Vector Fields Along Paths . . . . . . . . . . . . . . . 239

31 Integrals Over Surfaces 244

31.1 Regular Regions and Boundary Orientation . . . . . . . . . . . . 24431.2 Parameterized Regular Surfaces and Normals . . . . . . . . . . . 24531.3 Oriented Surfaces with Corners . . . . . . . . . . . . . . . . . . . 25031.4 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25431.5 Scalar Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . 25631.6 Surface Flux Integrals . . . . . . . . . . . . . . . . . . . . . . . . 25731.7 Generalized (n 1)-Dimensional Surfaces . . . . . . . . . . . . . 260

IV The Fundamental Theorems of Vector Calculus 263

32 Introduction to the Fundamental Theorem of Calculus 264

33 Greens Theorem in the Plane 266

34 Fundamental Theorem of Gradients 274

35 Stokes Theorem 277

36 The Divergence Theorem 282

37 Integration by Parts 288

38 Conservative vector fields 291

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Chapter 1

Introduction

This book is about the calculus of functions whose domain or range or both arevector-valued rather than real-valued. Of course, this subject is much too bigto be covered completely in a single book. The full scope of the topic containsat least all of ordinary differential equations, partial differential equation, anddifferential geometry. The physical applications include thermodynamics, fluidmechanics, elasticity, electromagnetism, and cosmology. Since a comprehensivetreatment is so ambitious, and since few undergraduates devote more than asemester to direct study of this subject, this book focuses on a much morelimited goal. The book will try to develop a series of definitions and results thatare parallel to those in an elementary course in the calculus of functions of asingle variable.

Consider the following syllabus for an elementary calculus course.

1. Precalculus

The arithmetic and algebra of real numbers. The geometry of lines in the plane: slopes, intercepts, intersections,

angles, trigonometry.

The concept of a function whose domain and range are both realnumbers and whose graphs are curves in the plane.

The concepts of limit and continuity2. The Derivative

The definition of the derivative as the limit of the slopes of secantlines of a function.

The interpretation of the derivative as the slope of the tangent line. The characterization of the tangent line as the best linear approxi-

mation of a differentiable function.

The development of various differentiation rules for products, com-posites, and other combinations of functions.

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For functions with a higher-dimensional domain, there are many waysto think of the derivative.

3. The Integral

We will consider several types of domains over which we will integratefunctions: curves, surfaces, oddly shaped regions in space.

4. The Fundamental Theorem of Calculus

We will find a whole hierarchy of generalizations of the fundamentaltheorem.

Our general procedure will be to follow the path of an elementary calculus courseand focus on what changes and what stays the same as we change the domainand range of the functions we consider.

Remark 1.1 (On notation). A wise man once said that, The more importanta mathematical subject is, the more versions of notation will be used for thatsubject. If the converse of that statement is true, vector calculus must beextremely important. There are many notational schemes for vector calculus.They are used by different groups of mathematicians and in different applicationareas. There is no real hope that their use will be standardized in the nearfuture. This text will use a variety of notations and will use different notationsin different contexts. I will try to be clear about this, but learning how to readand interpret the different notations should be an important goal for studentsof this material.

Remark 1.2 (On prerequisites). Readers are assumed to be familiar with thefollowing subjects.

Basic notions of algebra and very elementary set theory. Integral and differential calculus of a single variable. Linear algebra including solution of systems of linear equations, matrix

manipulation, eigenvalues and eigenvectors, and elementary vector spaceconcepts such as basis and dimension.

Elementary ordinary differential equations. Elementary calculations on real-valued functions of two or three variables

such as partial differentiation, integration, and basic graphing.

Of course, a number of these subjects are reviewed extensively, and I am mindfulof the fact that one of the most important goals of any course is to help thestudent to finally understand the material that was covered in the previouscourse. This study of vector calculus is a great opportunity to gain proficiencyand greater insight into the subjects listed above.

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Remark 1.3(On proofs). This text is intended for use by mathematicians andother scientists and engineers. While the primary focus will be on the calculation

of various quantities related to the subject, some effort will be made to providea rigorous background for those calculations, particularly in those cases wherethe proofs reveal underlying structure. Indeed, many of the calculations in thissubject can seem like nothing more than complicated recipes if one doesntmake an attempt to understand the theory behind them. On the other hand,this subject is full of places where the proofs of general theorems are technicalnightmares that reveal little (at least to me), and this type of proof will beavoided.

Remark 1.4 (On reading this book). My intention in writing this book is toprovide a fairly terse treatment of the subject that can realistically be read cover-to-cover in the course of a busy semester. Ive tried to cut down on extraneousdetail. However, while most of the exposition is directly aimed at solving the

problems directly posed in the text, there are a number of discussions that areintended to give the reader a glimpse into subjects that will open up in latercourses and texts. (Presenting a student with interesting ideas that he or shewont quite understand is another important goal of any course.) Many of theseideas are presented in the problems. I encourage students to read even thoseproblems that they have not been assigned as homework.

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Part I

Precalculus of Several

Variables

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Chapter 2

Vectors, Points, Norm, and

Dot Product

In this part of the book we study material analogous to that studied in a typi-cal precalculus course. While these courses cover some topics like functions,limits, and continuity that are closely tied to the study of calculus, the mostimportant part of such a course is probably the broader topic of algebra. Thatis true in this course as well, but with an added complication. Since we will bedealing with multidimensional objects vectors we spend a great deal of timediscussing linear algebra. We cover only relatively elementary aspects of thissubject, and the reader is assumed to be somewhat familiar with them.

Definition 2.1. We define a vector v Rn to be an n-tuple of real numbersv= (v1, v2, . . . , vn),

and refer to the numbers vi, i = 1, . . . , nas the componentsof the vector.We define two operations on the set of vectors: scalar multiplication

cv= c(v1, v2, . . . , vn) = (cv1, cv2, . . . , c vn)

for any real number c Rand vector v Rn, and vector addition

v + w= (v1, v2, . . . , vn) + (w1, w2, . . . , wn) = (v1+ w1, v2+ w2, . . . , vn+ wn)

for any pair of vectors v

Rn and w

Rn.

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Definition 2.2. If we have a collection of vectors

{v1, v2, . . . vk

}and scalars

{c1, c2, . . . , ck} we refer toc1v1+ c2v2+ + ckvk

as a linear combination of the vectors{v1, v2, . . . vk}.

Remark 2.3. We typically use boldface, lowercase, Latin letters to representabstract vectors in Rn. Another fairly common notation represents a vector by ageneric component with a free index, a subscript (usually, i,j , ork) assumedto range over the values from 1 to n. In this scheme, the vector v would bedenoted by vi, the vector x by xi, etc.

Remark 2.4. At this point we make no distinction between vectors displayedas columns or rows. In most cases, the choice of visual display is merely a matterof convenience. Of course, when we involve vectors in matrix multiplication thedistinction will be important, and we adopt a standard in that context.

Definition 2.5. We say that two vectors are parallel if one is a scalarmultiple of the other. That is, x is parallel to y if there exists c R suchthat

x= cy.

Remark 2.6. At this juncture, we have given the space of vectors Rn

only analgebraic structure. We can add a geometric structure by choosing an originand a set ofnperpendicular Cartesian axes forn-dimensional geometric space.With these choices made, every point X can be represented uniquely by itsCartesiancoordinates (x1, x2, . . . , xn). We then associate with every orderedpair of points X= (x1, x2, . . . , xn) andY = (y1, y2, . . . , yn) the vector

XY = (y1 x1, y2 x2, . . . , yn xn).

We think of this vector as a directed line segment or arrow pointing from the tailatXto the head atY. Note that a vector can be moved by parallel transportso that its tail is anywhere in space. For example, the vectorv= (1, 1) can berepresented as a line segment with its tail at X= (3, 4) and head at Y = (4, 5)

or with tail at X = (5, 7) and head at Y = (4, 8).This geometric structure makes vector addition and subtraction quite inter-esting. Figure 2.1 presents a parallelogram with sides formed by the vectorsxand y. The diagonals of the parallelogram represent the sum and difference ofthese vectors.

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a linear combination of the standard basis vectors

v=

ni=1

viei.

Definition 2.9. We define the (Euclidean) norm indexnorm of a vectorxRn to be

x=

x21+ x22+ + x2n=

ni=1

x2i .

A vector e is called a unit vectorife= 1.The distance between points X and Y (corresponding to the vectors x

andy) is given byXY=y x.

The dot product (or inner product) of two vectorsx Rn andy Rn isgiven by

x y= x1y1+ x2y2+ + xnyn=n

i=1

xiyi.

Remark 2.10. Note that for any nonzero vector v we can find a unit vector eparallel to that vector by defining

e= v

v.

In doing this we say we have normalized v.

Remark 2.11. The standard basis vectors have an important relation to thedot product.

vi = v ei.Thus, for any vector v

v=n

i=1

(v ei)ei.

Let us now note a few important properties of the dot product

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Theorem 2.12. For allx, y, w Rn and everyc R we have the following.1. (x + y) w= x w + y w.2. c(x y) = (cx) y= x (cy).3. x y= y x.4. x x0.5. x x= 0 if and only ifx= 0 = (0, 0, . . . , 0).

These are easy to prove directly from the formula for the dot product, and weleave the proof to the reader. (See Problem 2.8.)

Of course, there is an obvious relation between the norm and the dot product

x= x x. (2.1)However, we now prove a more subtle and interesting relationship.

Theorem 2.13 (Cauchy-Schwartz inequality). For allx, y Rn

|x y| xy.

Proof. For any real numberz R we compute0 x zy2

= (x zy) (x zy)= x (x zy) zy (x zy)= x x zx y zy x + z2y y= x2 2z(x y) + z2y2.

We note that quantity on the final line is a quadratic polynomial in the variablez. (It has the formaz2 +bz + c.) Since the polynomial is never negative, itsdiscriminant (b24ac) must not be positive (or else there would be two distinctreal roots of the polynomial). Thus,

(2x y)2 4x2y2 0,or

(x y)2 x2y2.Taking the square root of both sides and using the fact that|a|=

a2 for any

real number gives us the Cauchy-Schwartz inequality.

We now note that the norm has the following important properties

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Theorem 2.14. For allx, y Rn and everyc R we have the following.1.x 0.2.x= 0 if and only ifx= 0 = (0, 0, . . . , 0).3.cx=|c|x.4.x + y x + y (The triangle inequality).

Proof. One can prove the first three properties directly from the formula forthe norm. These are left to the reader in Problem 2.9. To prove the triangleinequality we use the Cauchy-Schwartz inequality and note that

x + y2 = (x + y) (x + y)= x (x + y) + y (x + y)= x x + x y + y x + y y= x2 + 2x y + y2 x2 + 2|x y| + y2 x2 + 2xy + y2= (x + y)2.

Taking the square root of both sides gives us the result.

Remark 2.15. While some of the proofs above have relied heavily on thespecific formulas for the norm and dot product, these theorems hold for moreabstract norms and inner products. (See Problem 2.10.) Such concepts are

useful in working with (for instance) spaces of functions in partial differentialequations where a common inner product between two functions defined onthe domain is given by the formula

f, g=

f(x)g(x)dx.

We will not be working with general inner products in this course, but it isworth noting that the concepts of the dot product and norm can be extended tomore general objects and that these extensions are very useful in applications.

Problems

Problem 2.1. Let x = (2, 5, 1), y = (4, 0, 8), and z = (1, 6, 7).(a) Compute x + y.(b) Compute z x.(c) Compute 5x.

(d) Compute 3z + 6y.

(a) Compute 4x 2y + 3z.

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Problem 2.2. Let x = (1, 3, 1), y = (2, 1, 3), and z = (5, 1, 2).(a) Compute x + y.

(b) Compute z x.(c) Compute3x.(d) Compute 4z 2y.(a) Compute x + 4y 5z.Problem 2.3. For the following two-dimensional vectors, create a graph thatrepresentsx, y,x,y, x y, and y x.(a) x = (2, 1), y = (1, 4).(b) x = (0, 3), y = (3, 4).(c) x = (4, 2), y = (5, 6).

Problem 2.4. For the points X and Y below compute the vectorsXY andY X.

(a)X= (4, 2, 6), Y = (2, 3, 1).(b)X= (0, 1, 4), Y = (3, 6, 9).(c)X= (5, 0, 5), Y = (1, 2, 1).

Problem 2.5. Let x = (1, 2, 0) and z = (1, 4, 3).(a) Computex.(b) Computez x.(c) Computez x.(d) Compute x z.

(e) Compute

x

z

z2 z.Problem 2.6. Let x = (2, 0, 1) and y = (1, 3, 2).(a) Computex.(b) Computey x.(c) Computey x.(d) Compute x y.(e) Compute

x yy2 y.

Problem 2.7. Use graphs of generic vectors x, y and x + y in the plane toexplain how the triangle inequality gets its name. Show geometrically the case

where equality is obtained.Problem 2.8. Use the formula for the dot product of vectors in Rn to proveTheorem 2.12.

Problem 2.9. Use the formula for the norm of a vector in Rn to prove the firstthree parts of Theorem 2.14.

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Problem 2.10. Instead of using the formula for the norm of a vector in Rn, use(2.1) and the properties of the dot product given in Theorem 2.12 to prove the

first three parts of Theorem 2.14. (Note that the proofs of the Cauchy Schwartzinequality and the triangle inequality depended only on Theorem 2.12, not thespecific formulas for the norm or dot product.)

Problem 2.11. Show that

x + y2 =x2 + y2

if and only ifx y= 0.

Problem 2.12. (a) Prove that ifx y= 0 for every y Rn then x = 0.(b) Prove that ifu y= v y for every y Rn thenu = v.Problem 2.13. The idea of a norm can be generalized beyond the particularcase of the Euclidean norm defined above. In more general treatments, anyfunction on a vector space satisfying the four properties of Theorem 2.14 isreferred to as a norm. Show that the following two functions on Rn satisfy thefour properties and are therefore norms.

x1 =|x1| + |x2| + + |xn|.

x= maxi=1,2,...,n

|xi|.

Problem 2.14. In R2 graph the three sets

S1 =

{x= (x, y)

R2

| x

1

},

S2 = {x= (x, y) R2 | x11},S3 = {x= (x, y) R2 | x1}.

Here is the Euclidean norm and 1and are defined in Problem 2.13.Problem 2.15. Show that there are constants c1, C1, c and C such thatfor every xRn

c1x1 x C1x1,cx x Cx.

We say that pairs of norms satisfying this type of relationship are equivalent.

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Chapter 3

Angles and Projections

While angle is a natural concept in R2 or R3, it is much harder to visualize inhigher dimensions. In Problem 3.5, the reader is asked to use the law of cosinesfrom trigonometry to show that ifx and y are in the plane (R2) then

x y=xy cos .

In light of this, we define the angle between two general vectors in Rn byextending the formula above in the following way. We note that ifxand y areboth nonzero then the Cauchy-Schwartz inequality gives us

|x y|xy1,

or 1 x yxy 1.

This tells us that xyxy is in the domain of the inverse cosine function, so wedefine

= cos1

x yxy

[0, ]

to be the angle between x and y. This gives us

cos = x yxy .

We state this definition formally and generalize the concept of perpendicularvectors in the following.

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Definition 3.1. For any two nonzero vectors x and y in Rn we define the

angle between the two vectors by

= cos1

x yxy

[0, ]

We say that x and y are orthogonal if x y = 0. A set of vectors{v1, v2, . . . , vk}is said to be an orthogonal setif

vi vj = 0 ifi=j.

We say that as set of vectors {w1, w2, . . . , wk} is orthonormal if it isorthogonal and each vector in the set is a unit vector. That is

wi wj =ij = 0 ifi=j,1 ifi = j.Example 3.2. The standard basis ei is an example of an orthonormal set.

Example 3.3. The set{(1, 1), (1, 1)}

is an orthogonal set in R2. The set

{(1/

2, 1/

2), (1/

2, 1/

2)}is an orthonormal set in R2.

The following computation is often useful

Definition 3.4. Let yRn be nonzero. For any vectorxRn we definethe orthogonal projection ofx onto y by

py(x) = x yy2 y.

The projection has the following properties. (See Figure 3.1.)

Lemma 3.5. For anyy=0 andx inRn we have

1. py(x) is parallel to y,2. py(x) is orthogonal to x py(x).The first assertion follows directly from the definition of parallel vectors.

The second can be shown by direct computation and is left to the reader. (SeeProblem 3.8.)

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y

x

py(x)

x py(x)

Figure 3.1: Orthogonal projection.

Example 3.6. Letx = (1, 2, 1) andy = (4, 0, 3). Thenx y= 1 andy= 5,so

py(x) = 1

25(4, 0, 3).

Note that

x py(x) =

21

25, 2,

28

25

and that (since py(x) and y are parallel)

py(

x) (

x

py(

x)) =

y (

x

py(

x)) = 0.

Problems

Problem 3.1. Compute the angle between the following pairs of vectors.

(a) x = (1, 0, 1, 1), y = (2, 2, 1, 0).(b) x = (3, 0, 1, 0, 1), y = (1, 1, 2, 1, 0).(c) x = (1, 0, 1), y = (5, 1, 0).Problem 3.2. Let x = (1, 2, 0), y = (3, 0, 1), and z = (1, 4, 3).(a) Compute py(x).

(b) Compute px(y).

(c) Compute py(z).

(d) Compute pz(x).

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Problem 3.3. Determine whether each of the following is an orthogonal set(a)

0001

,

1100

,

0011

.(b)

00

11

,

11

11

,

2011

.

(c)

1

00

1 ,

1101

,0010

.Problem 3.4. Determine whether each of the following is an orthonormal set(a)

1323023

,

23130

23

,

23

231313

.

(b)

12

0

12 ,

1313

13

,

00

1 .

(c)

12

12

0

,

161626

,

1313

13

.

Problem 3.5. Ifx and yare any two vectors in the plane, andis the (smallest)angle between them, the law of cosines1 from trigonometry says

x y2 =x2 + y2 2xy cos .

Use this to derive the identity

x y=xy cos .1Note that the law of cosines reduces to the Pythagorean theorem if = /2

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Problem 3.6. Suppose{w1, w2, . . . , wn}is an orthonormal set. Suppose thatfor some constants c1, c2, . . . , cn we have

x= c1w1+ c2w2+ + cnwn.

Show that for any i = 1, 2, . . . , n

x wi= ci.

Problem 3.7. Let{w1, w2, . . . , wk}be an orthonormal set in Rn and let xRn. Show that

ki=1

(x wi)2 x2.

Hint: use the fact that

0x

ki=1

(x wi)wi2

.

Problem 3.8. Show that for any vectors y= 0 and x in Rn the projectionpy(x) is orthogonal to x py(x).Problem 3.9. Show that for any x and nonzero y in Rn

py(py(x)) = py(x),

That is, the projection operator applied twice is the same as the projectionoperator applied once.

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Chapter 4

Matrix Algebra

In this section we define the most basic notations and computations involvingmatrices.

Definition 4.1. Anmn(read mbyn)matrixAis a rectangular arrayofmn numbers arranged in m rows and n columns.

A=

a11 a12 a1na21 a22 a2n

......

. . . ...

am1 am2 amn

.

We call ai1 ai2 ain

theith rowofA, (1im), and we call

a1ja2j

...amj

the jth column ofA, (1j n). We call the number aij in the ith rowand the jth column the ijth entry of the matrix A. The terms elementandcomponent are also used instead of entry.An abstract matrixA is often denoted by a typical entry

aij

with two free indices i (assumed to range from 1 to m) and j (assumed torange from 1 to n).

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Remark 4.2. The entries in matrices are assumed to be real numbers in thistext. Complex entries are considered in more complete treatments and will be

mentioned briefly in our treatment of eigenvalues.

Definition 4.3. As with vectors in Rn, we can define scalar multiplica-tion of any number c with an m n matrix A.

cA= c

a11 a12 a1na21 a22 a2n

......

. . . ...

am1 am2 amn

=

ca11 ca12 ca1nca21 ca22 ca2n

......

. . . ...

cam1 cam2 camn

.

Theij th entry ofcA is given by

c aij .

We can also define matrix addition provided the matrices have the samenumber of rows and the same number of columns. As with vector additionwe simply add corresponding entries

A + B =

a11 a12 a1na21 a22 a2n

......

. . . ...

am1 am2 amn

+

b11 b12 b1nb21 b22 b2n

......

. . . ...

bm1 bm2 bmn

=

a11

+ b11

a12

+ b12

a1n

+ b1na21+ b21 a22+ b22 a2n+ b2n

......

. . . ...

am1+ bm1 am2+ bm2 amn+ bmn

.

Theij th entry ofA + B is given by

aij+ bij.

Remark 4.4. Matrix addition is clearly commutative as defined, i.e.

A + B = B + A.

We define scalar multiplication to be commutative as well:

Ac= cA.

Example 4.5. Let

A=

3 2 41 5 7

,

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B= 0 1 32 8

9 ,

C=

6 04 7

.

Then

A + B=

3 + 0 2 + 1 4 31 + 2 5 + 8 7 9

=

3 1 11 13 2

,

and

2C=

2(6) 2(0)2(4) 2(7)

=

12 0

8 14

.

The sum A + Cis not well defined since the dimensions of the matrices do notmatch.

Definition 4.6. IfA is an m pmatrix andB is ap nmatrix, then thematrix product AB is an m n matrixCwhoseij th entry is given by

cij =

pk=1

aikbkj .

Remark 4.7. We note that theijth entry ofAB is computed by taking the ith

row ofA and the j th row ofB (which we have required to be the same length(p)). We multiply the rows term by term and add the products (as we do intaking the dot product of two vectors).

a11 a12 . . . a1p

......

... ai1 ai2 . . . a1p

am1 am2 . . . amp

b11 b21

...bp1

b1jb2j

...bpj

b1n b2n

... bpn

=

c11 c1j c1nc21 c2j c2n

......

...ci1 [cij] cin

... ... ...cm1 cmj cmn

.

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Example 4.8.

6 04 7

0 1 32 8 9

=

6(0) + 0(2) 6(1) + 0(8) 6(3) + 0(9)4(0) 7(2) 4(1) 7(8) 4(3) 7(9)

=

0 6 1814 52 51

.

Remark 4.9. Matrix multiplication is associative. That is,

A(BC) = (AB)C

whenever the dimensions of the matrices match appropriately. This is easiest todemonstrate using component notation. We use the associative law for multi-plication of numbers and the fact that finite sums can be taking any order (thecommutative law of addition) to get the following.

nj=1

aij(m

k=1

bjkckl) =n

j=1

mk=1

aijbjkckl =m

k=1

(n

j=1

aijbjk)ckl.

Remark 4.10. Matrix multiplication is not commutative. To compute ABwe must match the number of columns on the left matrix A with the numberof rows of the right matrix B . The matrix B A. . .

might not be defined at all,

might be defined but of a different size than AB , or

might be have the same size as AB but have different entries.We will see examples of this in Problem 4.2 below.

Remark 4.11. When a vector xRn is being used in matrix multiplication,we will always regard it as a column vector or n 1 matrix.

x=

x1x2...

xn

.

Definition 4.12. IfA is an n nmatrix, the elements aii, i = 1, . . . , narecalled the diagonal elements of the matrix. A is said to be diagonal ifaij = 0 whenever i=j .

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Definition 4.13. The n

n diagonal matrix I with all diagonal elements

aii = 1, i = 1, . . . , nis called the n n identity matrix.

I=

1 0 00 1 0...

... . . .

...0 0 1

.

The entries of the identity can be represented by the Kronecker delta func-tion,ij.

The identity matrix is a multiplicative identity. For any m nmatrixB wehave

IB = BI=B.

Note that here we are using the same symbol to denote the mm and then n identity in the first and second instances respectively. This ambiguity inour notation rarely causes problems. In component form, this equation can bewritten

mk=1

ikbkj =n

k=1

bikkj =bij.

Definition 4.14. We say that an nn matrix A is invertible if thereexists a matrix A1 such that

AA1 =A1A= I .

Example 4.15. Suppose that

A=

a b

c d

andad bc= 0. Then one can check directly that

A1 = 1

ad bc

d bc a

.

Lemma 4.16. A matrix has at most one inverse.The proof of this is left to the reader in Problem 4.3.

Lemma 4.17. SupposeA andB are invertiblen n matrices. Then(AB)1 =B1A1.

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The proof of this is left to the reader in Problem 4.4.

Definition 4.18. Thetransposeof anmnmatrixA is thenmmatrixAT obtained by using the rows ofA as the columns ofAT. That is, theij th

entry ofAT is the j ith entry ofA.

aTij =aji.

We say that a matrix is symmetric ifA = AT andskew ifA= AT.

Example 4.19.

0 1 32 8 9

T

= 0 21 8

3 9 .The matrix

2 1 01 2 10 1 2

is symmetric. The matrix 0 2 52 0 4

5 4 0

is skew.

The next lemma follows immediately from the definition.

Lemma 4.20. For any matricesA andB and scalarc we have

1. (AT)T =A.

2. (A + B)T =AT + BT ifA andB are bothm n.3. (cA)T =c(AT).

4. (AB)T =BTAT ifA ism p andB isp n.5. (A1)T = (AT)1 ifA is an invertiblen n matrix.The proof of this is left to the reader in Problem 4.5. We also note the

following.

Lemma 4.21. IfA is anm n matrix, x Rm, andy Rn then

x (Ay) = (ATx) y.

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Proof. If we look at this equation in component form we see that it followsdirectly from the definition of multiplication by the transpose and the associative

and commutative laws for multiplication of numbersn

i=1

xi(

mj=1

aijyj) =

ni=1

mj=1

xiaijyj =

mj=1

(

ni=1

xiaij)yj .

Definition 4.22. An n n matrix Q is orthogonal if

QQT =QTQ= I .

That is, ifQT =Q1.

Example 4.23. The 2 2 matrix cos sin

sin cos

is orthogonal since cos sin

sin cos

cos sin sin cos

=

cos sin sin cos

cos sin

sin cos

=

cos2 + sin2 0

0 cos2 + sin2

=I .

Problems

Problem 4.1. Let

A=

2 34 1

, B=

0 71 5

,

C=

6 17 8

2 4

, D=

9 3 00 4 7 .

(a) Compute 2A.

(b) Compute 4A 2B.(c) Compute C 3DT.(d) Compute 2CT + 5D.

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Problem 4.2. Let

A=

2 34 1

, B=

0 71 5

,

C=

6 17 8

2 4

, D= 9 3 0

0 4 7

.

(a) ComputeAB .

(b) ComputeB A.

(c) Compute C D.

(d) ComputeDC

Problem 4.3. Show that the inverse of an n

n matrix A is unique. That is,

show that ifAB= BA= AC= C A= I ,

thenB = C.

Problem 4.4. Show that ifA and B are invertiblen n matrices then

(AB)1 =B1A1.

Problem 4.5. Prove Lemma 4.20.

Problem 4.6. Show that every n nmatrixA can be written uniquely as thesum of a symmetric matrix Eand a skew matrix W. Hint: IfA = E+ W thenAT =? We refer to Eas the symmetric part ofA and Was the skew part.

Problem 4.7. While we dont use it in this text, there is a natural extensionof the dot product for n nmatrices:

A, B=n

i=1

nj=1

aijbij .

Show that ifA is symmetric and B is skew thenA, B= 0.Problem 4.8. Let A be any nn matrix and let E be its symmetric part.Show that for any x Rn we have

xTAx= xTEx.

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Chapter 5

Systems of Linear

Equations and Gaussian

Elimination

One of the most basic problems in linear algebra is the solution of a system mlinear equations in n unknown variables. In this section we give a quick reviewof the method of Gaussian elimination for solving these systems.

The following is a generic linear system.

a11x1+ a12x2+ + a1nxn = b1,a21x1+ a22x2+ + a2nxn = b2,

.

..am1x1+ am2x2+ + amnxn = bm.

Here, we assume that aij , i = 1, . . . , m, j = 1, . . . , n and bi, i = 1, . . . , m areknown constants. We call the constants aij the coefficients of the system.The constants bi are sometimes referred to as the data of the system. Thenvariables xj ,j = 1, . . . , nare called the unknownsof the system. Any orderedn-tuple (x1, x2, . . . , xn) Rn that satisfies each of themequations in the systemsimultaneously is a solution of the system.

We note that the generic system above can be written in term of matrixmultiplication.

a11 a12 a1na21 a22 a2n... ... . . . ...am1 am2 amn

x1

x2...

xn

=

b1

b2...

bm

,

orAx= b.

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Here A is the m n coefficient matrix, xRn is the vector of unknowns, andb

Rm is the data vector.

It is worth considering the very simple case where n = m = 1. Our equationreduces to

ax= b

where a, x, and b are all real numbers. (We think of a and b as given; x isunknown.) The only alternatives for solutions of this equation are as follows.

Ifa= 0 then the equation has the unique solution x = ba . Ifa = 0 then there are two possibilities.

Ifb = 0 then the equation 0 x= 0 is satisfied by any xR. Ifb= 0 then there is no solution.

We will see these three alternatives reflected in our subsequent results, but wecan get at least some information about an important special case immediately.We call a system ofm equations inn unknowns (or the equivalent matrix equa-tionsAx= b) homogeneous ifbi = 0, i = 1, . . . , m, (or equivalently, b= 0).We note that every homogeneous system has at least one solution, the trivialsolutionxj = 0, j = 1, . . . , n, (x= 0).

More generally, a systematic development of the method of Gaussian elim-ination (which we wont attempt in this quick review) reveals an importantresult.

Theorem 5.1. For any linear system ofm linear equations inn unknowns,exactly one of the three alternatives holds.

1. The system has a unique solution(x1, . . . , xn).

2. The system has an infinite family of solutions.

3. The system has no solution.

The following examples of the three alternatives are simple enough to solve byinspection or by solving the first equation for one variable and substituting thatinto the second. The reader should do so and verify the following.

Example 5.2. The system

2x1 3x2 = 1,4x1+ 5x2 = 13,

has only one solution: (x1, x2) = (2, 1).

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Example 5.6. Multiplying the second row of

0 3 12 6 103 7 9

by 12 yields

0 3 11 3 53 7 9

.

Interchanging the first two rows of this matrix yields

1 3 50 3 1

3 7 9

.

Adding three times the first row to the third of this matrix yields 1 3 50 3 1

0 2 24

.

Multiplying the third row of this matrix by12 yields 1 3 50 3 1

0 1 12

.

Interchanging the second and third rows yields 1 3 50 1 12

0 3 1

.

Adding3 times the second row of this matrix to the third yields 1 3 50 1 12

0 0 35

.

Dividing the final row by 35 yields

1 3 50 1 120 0 1

.Elementary row operations have an important property: they dont change

the solution set of the equivalent systems.

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Theorem 5.7. Suppose the matrixB is obtained from the matrix A by a

sequence of elementary row operations. Then the linear system representedby the matrixB has exactly the same set of solution as the system representedby matrixA.

The proof is left to the reader. We note that it is obvious that interchangingequations or multiplying both sides of an equation by a nonzero constant (whichare equivalent to the first two operations) doesnt change the set of solutions.It is less obvious that the third type of operation doesnt change the set. Is iteasier to show that the operation doesnt destroy solutions or doesnt create newsolutions? Can you undo a row operation of the third type by doing anotherrow operation?

Example 5.8. If we interpret the matrices in Example 5.6 as augmented ma-

trices, the first matrix represents the system

3x2 = 1,2x1 6x2 = 10,

3x1+ 7x2 = 9.

while the final matrix represents the system

x1 3x2 = 5,x2 = 12,

0 = 1.

According to our theorem these systems have exactly the same solution set.While it is not that hard to see that the first system has no solutions, theconclusion is immediate for the second system. That is because we have usedthe elementary row operations to reduce the matrix to a particularly convenientform which we now describe.

Gaussian elimination is the process of using a sequence of elementary rowoperations to reduce an augmented matrix in a standard form called reducedrow echelon formfrom which it is easy to read the set of solutions. The formhas the following properties.

1. Every row is either a row of zeros or has a one as its first nonzero entry(a leading one).

2. Any row of zeros lies below all nonzero rows.

3. Any column containing a leading one contains no other nonzero entries.

4. The leading one in any row must lie to the left of any leading one in therows below it.

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Example 5.9. The matrix

1 0 0 30 1 0 20 0 1 70 0 0 00 0 0 0

is in reduced row echelon form. If we interpret the matrix as an augmentedmatrix corresponding to a system of five equations in three unknowns it isequivalent to the matrix equation

1 0 00 1 00 0 1

0 0 00 0 0

x1x2

x3

=

327

00

,

or the system of five scalar equation

x1 = 3,

x2 = 2,

x3 = 7,

0 = 0,

0 = 0.

Of course, the unique solution is (x1, x2, x3) = (3, 2, 7).

Example 5.10. Again, the matrix

1 0 2 50 1 4 60 0 0 00 0 0 00 0 0 0

is in reduced row echelon form. If we again interpret the matrix as an augmentedmatrix corresponding to a system of five equations in three unknowns it isequivalent to the matrix equation

1 0 2

0 1 40 0 00 0 00 0 0

x1x2

x3

= 5

6000

,

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Problem 5.1 gives a number of examples of matrices in reduced row echelonform and asks the reader to give the set of solutions. There is an intermediate

form called row echelon form. In this form, columns are allowed to havenonzero entries above the leading ones (though still not below). From this formit is easy to determine which of the three solution alternatives hold. Problem 5.2gives a number of examples of matrices in this form. The reader is asked todetermine the alternative by inspection and then determine all solutions wherethey exist.

Since this is a review, we will not give an elaborate algorithm for usingelementary row operation to reduce an arbitrary matrix to reduced row echelonform. (More information on this is given in the references.) We will contentourselves with the following simple examples.

Example 5.12. The system in Example 5.2 can be represented by the aug-mented matrix

2 3 14 5 13

.

In order to clear out the first column, let us add2 times the first row to thesecond to get

2 3 10 11 11

.

We now divide the second row by 11 to get 2 3 10 1 1

.

Adding 3 times the second row to the first gives 2 0 40 1 1

.

Finally, dividing the first row by 2 puts the matrix in reduced row echelon form 1 0 20 1 1

.

This is equivalent to the system

x1 = 2,

x2 = 1,

which describes the unique solution.Note that we chose the order of our row operation in order to avoid introduc-

ing fractions. A computer computation would take a more systematic approachand treat all coefficients as floating point numbers, but our approach makessense for hand computations.

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Example 5.13. The system in Example 5.3 can be represented by the aug-mented matrix

1 1 52 2 10

.

Taking2 times the first row of this matrix and adding it to the second in orderto clear out the first column yields

1 1 50 0 0

.

This is already in reduced row echelon form and is equivalent to the equation

x1 x2 = 5.

Since the second column has no leading one, we let the corresponding variable,

x2 take on an arbitrary value x2 = s R and solve for the system for thosevariables whose column contains a leading one. (In this case, x1) Our solutionscan be represented as

x1 = s + 5,

x2 = s,

for any s R.Example 5.14. The system in Example 5.4 can be represented by the aug-mented matrix

1 1 12 2 7

.

Taking2 times the first row of this matrix and adding it to the second in orderto clear out the first column yields

1 1 10 0 5

.

Without even row reducing further, we can see that the second row representsthe equation

0 = 5.

Therefore, this system can have no solutions.

Problems

Problem 5.1. The following matrices in reduced row echelon form representaugmented matrices of systems of linear equations. Find all solutions of thesystems.

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(a)

1 0 0

2

0 1 0 30 0 1 50 0 0 00 0 0 0

.

(b)

1 4 0 0 3 20 0 1 0 2 30 0 0 1 2 50 0 0 0 0 0

.

(c)

1 0 0 2 00 1 0 4 0

0 0 1 7 00 0 0 0 1

.Problem 5.2. The following matrices in row echelon form represent augmentedmatrices of systems of linear equations. Determine by inspection which of thethree alternatives hold: a unique solution, an infinite family of solutions, or nosolution. Find all solutions of the systems that have them.

(a)

1 2 2 2 20 1 5 3 20 0 1 7 60 0 0 0 1

.

(b)

1 4 0 2 3 20 1 1 4 2 30 0 0 1 2 50 0 0 0 1 2

.

(c) 1 1 4 20 1 3 3

0 0 1 5

.

Problem 5.3. The following matrices represent augmented matrices of systemsof linear equations. Find all solutions of the systems.

(a) 3 1 1 24 0 2 21 3 2 3

.(b)

3 6 1 6 02 4 3 7 0

.

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(c)

2 3 2 11 1 4 2 .Problem 5.4. Find all solutions of the following systems of equations.

(a)

x1+ x2+ x3 = 0,

x1 x2 = 0,2x1+ x3 = 0.

(b)

3u + 2v w = 7,

u v+ 7w = 5,2u + 3v 8w = 6.

(c)

6x 14y = 28,7x + 3y = 6.

(d)

c1 c2 = 6,c2 + c3 = 5,

c3

c4 =

4,

c1 c4 = 15.

(e)

2x + 7y+ 3z 2w = 8,x + 5y+ 3z 3w = 2,

2x + 3y z 2w = 4,3x y+ z+ 3w = 2,

2x + 2y+ 3z+ w = 1.

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Definition 6.7. For an n

n matrix we define

det

a11 a12 a1na21 a22 a2n

......

. . . ...

an1 an2 ann

=

a11 a12 a1na21 a22 a2n

......

. . . ...

an1 an2 ann

=

ni1=1

ni2=1

n

in=1

i1i2...ina1i1a2i2 anin .

For a 3 3 matrix this has the form

deta11 a12 a13a21 a22 a23

a31 a32 a33 =

a11 a12 a13a21 a22 a23

a31 a32 a33 =

3

i=1

3

j=1

3

k=1

ijka1ia2ja3k.

While this definition has the advantage of being explicit and simple to state,it is not an easy formula to compute. It has, after all, nn terms in the sum,which can get large very quickly. Even if we note that only n! of the termsare nonzero, this is still a large number for even a moderate size ofn. In orderto deal with this, we use our basic definition to derive a family of formulascalled expansion be cofactors. This will give us fairly reasonable methodcomputation of determinants, though as we will see, there is no getting aroundthe fact that in general, computing the determinant of a matrix is a laborintensive process.

Lets examine the formula for the determinant of a 3

3 matrix. Fortunately,

while the sum is over 33 = 27 terms, only 3! = 6 of them are nonzero.

det

a11 a12 a13a21 a22 a23

a31 a32 a33

= a11a22a33+ a12a23a31+ a13a21a32

a11a23a32 a13a22a31 a12a21a33.

A few observations will lead us to some important results. Note the following:

Each of the 3! nonzero terms in the sum has 3 factors. One factor from each term comes from each row.

One factor from each term comes from each column.

Now note that for each ordered pair ij , the entry aij appears in two (or 2!)terms of the sum. We can factor aij out of the sum of these two terms and writethose terms in the form

aijAij

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where we call Aij the cofactor ofaij . Since each term of the full determinanthas one factor from every row (and every column) of the original matrix, we can

write the determinant in terms of the factors from particular any row or anycolumn. That is, for any i = 1, 2, 3 orj = 1, 2, 3 we can write

det A = ai1Ai1+ ai2Ai2+ ai3Ai3

= a1jA1j+ a2jA2j+ a3jA3j .

At this point, let us observe that the observations above easily generalize tothen ncase.

Each of the n! nonzero terms in the determinant of an n n matrix hasn factors.

One factor from each term comes from each row.

One factor from each term comes from each column.

Since each term of the determinant has one factor from every row (and everycolumn) of the original matrix, we can write the determinant in terms of thefactors from either the ith row or thej th column and write the determinant of

det A = ai1Ai1+ ai2Ai2+ + ainAin= a1jA1j+ a2jA2j+ + anjAnj.

All of this would be pretty useless if there wasnt a convenient formula forthe cofactors of a matrix. Fortunately, one can show the following.

Lemma 6.8. For anyi, j = 1, 2, . . . , n

Aij = (1)i+jMij

whereMij is the ijth

minor of A the determinant of the (n 1) (n 1)matrix obtained by deleting the ith row andjth column of the matrixA.

The proof of this result in the general n n case requires more informationabout permutations than will be covered in this text. However, the result can beobtained by direct computation in the 33 case. For example, we can rearrangethe formula for the 3 3 determinant above, factoring out all terms from, say,the second row

det A = a11a22a33+ a12a23a31+ a13a21a32

a11a23a32 a13a22a31 a12a21a33= a21(a13a32 a12a33) + a22(a11a33 a13a31) + a23(a12a31 a11a32)= (

1)2+1a21 a12 a13

a32 a33 + (1)

2+2a22 a11 a13a31 a33

+(1)2+3a23

a11 a12a31 a32 .

We call this technique for computing the determinant expansion by co-factorsand we state it now as a theorem.

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The following theorem is quite important.

Theorem 6.14. For anyn n matricesA andB we have the following.1.

det(AB) = det A det B.

2. If A is invertible thendet A= 0 and

det(A1) = 1

det A.

Unfortunately, the proof of the first part requires some techniques that are notcovered in this book. It is covered in most standard texts on linear algebra.

The second part follows directly from the first and the identity

1 = det I= det(AA1) = det A det A1.

Definition 6.15. An orthogonal matrix Q is called a rotation indexrota-tion matrix if det Q= 1.

Remark 6.16. Since for any orthogonal matrix we have

det(QQT) = (det Q)2 = det I= 1,

it follows that det Q=1. Thus, rotation matrices satisfy det Q= 1. The readershould verify that the orthogonal matrices in Example 4.23 are rotations.

While (as we will see in the rest of the book) determinants have many ge-ometric applications, one of the most important applications is to the problemn linear equations in n unknowns. The following theorem should be familiar tothe reader.

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Theorem 6.17. LetAbe annnmatrix. Then the following are equivalent.(That is, if one statement is true all are true. If one statement is false all are false.)

1. For everybRn the matrix equation

Ax= b

has a unique solution.

2.det A= 0.

3. The matrixA is invertible.

4. The matrixA can be row reduced to the identity matrix.5. The homogeneous system

Ax= 0

has only the trivial solution.

Again we direct the reader to standard linear algebra texts for the proof of thistheorem.

Problems

Problem 6.1. Compute the determinants of the following matrices.

A= 2 34 1 , B= 0 71 5 ,

C=

2 00 5

, D=

9 318 6

.


A=

1 2 32 4 6

5 4 1

, B=

2 0 70 3 5

0 0 5

,

C=

6 1 07 8 1

2 4 1

, D=

9 3 00 4 71 0 3

.


A=

2 5 0 12 0 3 2

0 7 0 61 4 1 1

, B =

3 1 2 00 4 1 20 0 0 60 0 1 1

.

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Chapter 7

The Cross Product and

Triple Product in R3

To this point we have defined two types of products involving vectors.

1. Thescalar productof a scalar c and a vector v Rn yields a vector

cvRn.

The resulting vector is parallel to the original vector.

2. Thedot product or inner productof two vectors u, v Rn yields a scalar

u vR.

The result gives us information about the angle between the two vectors.

In the special case ofR3 we define thecross productof two vectors which yieldsanother vector.

Definition 7.1. For any vectors u = (u1, u2, u3) and v = (v1, v2, v3) in R3

we define the cross product to be the vector

u v=3

i=1

3j=1

3k=1

ijkuivjek.

Remark 7.2. This equation can be expanded in the forms

u v = (u2v3 u3v2)e1+ (u3v1 u1v3)e2+ (u1v2 u2v1)e3= (u2v3 u3v2)i + (u3v1 u1v3)j + (u1v2 u2v1)k.

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Theorem 7.5. The cross product has the following properties.

1. For allu andv inR3

u v=v u.

2. For alluR3u u= 0.

3. For allu, v, andw inR3 and real numbers and

u (v + w) = (u v) + (u w),(u + v) w = (u w) + (v w).

4. For any i= 1, 2, 3 andj = 1, 2, 3

ei ej =3

k=1

ijkek.

This reduces to

e1 e2 = e3,e2 e3 = e1,e3 e1 = e2.


u (u v) = v (u v) = 0.


u v=uv| sin |,

where is the (smallest) angle between the vectorsu andv

Proof. Parts 1, 2, and 3 follow directly from the properties of the determinant(see Theorem 6.12) and the mnemonic for the cross product. Part 4 can beeasily verified through direct computation.

Part 5 can be obtained through a formula that will be of interest below. For

anyu, v, and w in R3 we note that

w (u v) =

w1 w2 w3u1 u2 u3v1 v2 v3

.This follows directly from the determinant mnemonic for the cross product and

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system in three dimensions. It is pretty easy to convince yourself that we canchoose any two perpendicular directions for the first two positive coordinate

axes corresponding to e1 and e2. However, once we have done this there aretwo ways to direct the third axis (or e3).

Definition 7.8. We call a coordinate system for R3 right-handedif push-ing e1 toward e2 with the palm of a typical right hand causes the thumbto point in the direction of e3. If e3 points in the opposite direction, thecoordinate system is called left-handed.

It is standard to use a right-handed coordinate system for R3. If this is thecase, the direction ofu vis determined by theright-hand rule: Ifu is pushedtowardv with the palm of your right hand, your thumb will point toward u

v.

In the proof of the previous theorem, we computed a quantity that has aninteresting physical interpretation.

Definition 7.9. For any u, v, and w in R3 we define the vector tripleproduct of the three vectors to be

w (u v) =

w1 w2 w3u1 u2 u3v1 v2 v3

.

Remark 7.10. We can see from the determinant formula that the order of thevectors matters only up to a sign. That is

w (u v) = u (v w) = v (w u)= u (w v) =w (v u) =v (u w).

We can compute the magnitude of this quantity

|w (u v)|= cos wu vwhere is the angle between w andu v. Note the following.

The quantitycos

w

gives us thealtitudeof the vectorw above the plane determinedu and v. The quantity

u vgives us the area of the parallelogram with sides u and v.

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The volume of a parallelepiped (or, more generally, any prism) is theproduct of the area of the base and the distance between the top and

bottom face (the altitude)

Putting these together gives us the following.

Theorem 7.11. The magnitude of the vector triple product of three vectorsis the volume of a parallelepiped with the three vectors as sides.

u

v

w

u v

cos w

Figure 7.2: Parallelepiped created by the triple of vectors u. v, and w.

Problems

Problem 7.1. Let u = (2, 1, 3), v = (4, 0, 1),w= (3, 1, 1).(a) Compute u v.(b) Compute w u.(c) Compute w (u v).

(d) Compute u(uv). (Note you should know the result before doing thecomputation.)Problem 7.2. Let u = i + 3j + k, v = 2j + 3k,w= i k.(a) Compute u v.(b) Compute w u.

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(c) Compute w (u v).(d) Compute u

(v

w). (Note you should know the result before doing the

computation.)

Problem 7.3. Find all unit vectors orthogonal to both the vectors given below.

(a) (1, 2, 1) and (3, 3, 4).(b) (2, 1, 5) and (1, 0, 2).

(c) j andk.(d) i 2kand j + k.Problem 7.4. Find the area of a parallelograms with sides given by the pairsof vectors in Problem 7.3

Problem 7.5. Find the area of the triangle with vertices (1, 0, 1), (2, 2, 2), and

(1, 2, 2).Problem 7.6. Find the volumes of the parallelepipeds with edges determinedby the vectors given below.

(a) (1, 0, 0), (1, 2, 1), and (3, 3, 4).(b) (1, 1, 3), (2, 1, 5), and (1, 0, 2).

(c) i k, j, andk.(d) e1+ e2, e1 2e3, and e2+ e3.Problem 7.7. Prove the identity

a (b c) = (a c)b (a b)c.

This is commonly known as the bac-cab rule. (You can write the right sideas b(a c) c(a b).)Problem 7.8. Is the cross product associative? In other words, is it alwaystrue that

a (b c) = (a b) c?If not, under what conditions on a, b, and c does the equation above hold?

Problem 7.9. Show that the absolute value of

det

u1 u2v1 v2

is the area of the parallelogram with sides given byu= (u1, u2) andv= (v1, v2).Hint: show that the square of the determinant isu2v2 sin2 .Problem 7.10. Find the area of the parallelogram with vertices (1, 3), (0, 4),(1, 2), (2, 3).

Problem 7.11. Show that ifu v= 0 then u and v are parallel.

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Example 8.4. To find the equation of a plane containing the points A =(3.

2, 5),B = (9,

3, 7) andC= (4, 6, 0), we first find two vectors in the plane.

For instance, we can take u =AB= (6, 1, 2) and v =BC= (5, 9, 7). Thecross product of these vectors is normal to the plane.

n= u v=

i j k

6 1 25 9 7

= 25i + 52j + 49k.UsingA as our reference point in the plane we get the equation

25(x 3) + 52(y+ 2) + 49(z 5) = 0.

Example 8.5 (Parametric representation of a plane). We note that the equa-tion for a planeax + by + cz = d can be solved by Gaussian elimination (though

there is nothing much to eliminate). Ifa= 0 the reduced row echelon form ofthe augmented matrix would be

1,b

a,c

a,d

a

.

We would assign arbitrary values y = s and z = t to the two variables withoutleading ones in their columns and solve for x to get the solutions

x = d

a s b

a t c

a,

y = s,

z = t,

for any s R and tR. This can be written in vector form as xy

z

=p(s, t) =

da0

0

+ s

ba1

0

+ t

ca0

1

.

Here p is a function from R2 to R3 defined for any values of the parameters sandt.

Example 8.6 (System of equations representing a line). How can we bestdescribe the line defined by the intersection of two planes. For example considerthe line defined by the two planes

x + y z = 4, and2x + 3y+ z = 9.

Of course, we really dont need to do anything more to describe the line. Thepoints (x, y, z) on the line are simply the solutions of the system of two equationsin three unknowns. However, if we wish to express the line in parametric form,

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we need to solve the system so that the solutions are described as a function.The system can be represented as the augmented matrix.

1 1 1 42 3 1 9

.

This reduces to the matrix 1 0 4 30 1 3 1

,

which yields the solutions

x = 4t + 3,

y = 3t + 1,z = t,

for alltR. Of course, we can write this as the equation of a line in parametricform

(x, y, z) = l(t) = (3 + 4t, 1 3t, t), tR.In Problem 8.9 we ask the reader to find the equations of two planes that

intersect in a line with a given parametric representation.

Problems

Problem 8.1. Find a parametric equation of the line containing the point(2, 7, 4) and parallel to the vector (3, 5, 0).Problem 8.2. Find a parametric equation of line containing the points (8, 0,

6)

and (7, 9, 3).

Problem 8.3. Find the equation of the plane containing the point (3, 2, 1) andnormal to the vector (1, 4, 0).Problem 8.4. Find the equation of the plane containing the line

(x,y,z) = l1(t) = (5 + 4t, 5 3t, 2), tR,and perpendicular to the line

(x,y,z) = l2(t) = (2 3t, 3 4t, 5 + 7t), tR.Problem 8.5. Do the lines

l1(t) = (5 + 3t, 1 + 7t, 4 2t), t R,and

l2(t) = (6 2t, 9 + 3t, 4 + 2t), tR,intersect? If so, where? If there is a plane containing the two lines find anequation for it.

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The boundary ofA has two pieces: the circle of radius one and the circle ofradius two. Every point x

A is an interior point. If we take = min

{2

x, x 1}/2 (that is, half the distance between x and the boundary) thenthe ball of radius about x lies inside the annulus. The closure ofAis given by

A={x R2 |1 x 2}.

Example 9.6. The punctured unit disk

D={xR2 |0 0 there exists >0 such that for any x with0


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While the function is not defined at the origin, we can ask if the limit exists as(x, y)

(0, 0). In this case we note that

f(xi, yi) = x4i y4i

x2i + y2i

= (x2i y2i )(x2i + y2i )

x2i + y2i

= x2i y2i 0 0 = 0 as (xi, yi)(0, 0).

So the limit exists and

lim(x,y)(0,0)

x4 y4x2 + y2

= 0.

The following theorem (which we state without proof) show that the limitsof various algebraic combinations of functions behave in the obvious way.

Theorem 9.9. Let Rm be the domain of functions f : Rn andg: Rn Suppose that for somex0 we have

limxx0

f(x) = lf,

limxx0

g(x) = lg.

Then,

1. limxx0cf(x) = clf for every scalarc R,

2. limxx0v f(x) = v lf for every vectorv Rn

,3. limxx0Af(x) = Alf for everyk n matrixA,4. limxx0(f(x) + g(x)) = lf+ lg,

5. limxx0(f(x) g(x)) = lf lg,6. limxx0(f(x) g(x)) = lf lg ifn= 3.

The next theorem allows us to compute the limit of a product of functionswhen the limit of one function is zero even if we know only that the secondfunction is bounded.

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Theorem 9.10. Let Rm be the domain of functions f : Rn andg: R Suppose that for somex0 we have

limxx0

f(x) = 0,

and thatg is bounded in a neighborhood ofx0. That is, there existsM >0and >0 such that

|g(x)| Mfor allx B(x0). Then

limxx0

g(x)f(x) = 0,

Proof. Let >0 be given. Since limxx0f(x) = 0 there exists 1 > 0 such thatfor every x with 0


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2. A function might not have a limit because it becomes unbounded. In thiscase we say it diverges to plus or minus infinity at a point. This is the

case withf(x) =

1

x2

which diverges to infinity as x approaches 0 from either the right or left.

3. A function would not have a limit if it oscillates so wildly that the limitdoes not exist. This is the case with the function

g(x) = sin

1

x

which has no limit as x0.An important key here is that in one dimension there are really only two paths

to a point: from the right or the left. When we move to two dimensions (orhigher) there are an infinitenumber of paths we can take to a point.

Example 9.13. The function

f(x, y) = x2

x2 + y2

has no limit as (x, y) approaches (0, 0). To see this, suppose we choose a sequenceof points along the line through the origin y = x where is any constant. Alongsuch a line we would have

f(x,x) = x2

x2 + 2x2 =

1

1 + 2.

Thus, the function is constant along any line through the origin, but the constantchanges depending on the direction we choose. To visualize the graph of thefunction, think of a spiral staircase, where the height as one goes toward thecenter post depends on which step you are on.

As it did in one-dimensional calculus, the notion of limit of a function leadsdirectly to the notion of continuity.

Definition 9.14. Let Rn. We say that the function f : Rm iscontinuous at x0 if

limxx0 f(x) = f(x0).

Iffis continuous at every x we say that fis continuous on and writef C().

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x

y

x2

x2 y2

x

y

Figure 9.1: Graph of the function f(x, y) = (x2)/(x2 +y2) which has a spiralstaircase discontinuity at the origin. Lines going into the origin have a differentlimiting height.

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Example 9.15. Our examples above of limits of functions with multidimen-sional domains can be interpreted in terms of continuity. For instance, the

function

f(x, y) =

x4y4x2+y2 , (x, y)= (0, 0)

0, (x, y) = (0, 0)

is continuous. Since the limit of the rational function x4y4

x2+y2 existed as (x, y)(0, 0), we could simply extend the domain to the origin and define the value ofthe extended function to be the limit. However, the function

g(x, y) =

x2

x2+y2 , (x, y)= (0, 0)0, (x, y) = (0, 0)

is not continuous. In fact, there is no way to extend the domain of the functionx2

x2+y2 to the origin since the limit of the function does not exist at that point.

The following theorem, (which we state without proof) says that variousalgebraic combinations of continuous functions are continuous.

Theorem 9.16. LetRn, and letcR.1. Iff : Rm is continuous atx0 then so is the constant multiple

cf.

2. If f : Rm and g : Rm are continuous at x0 then so isthe sum f+ g.

3. Iff :

R andg :

R are continuous at x0

then so is theproductf g.

4. Iff : R andg : R are continuous atx0 andg(x0)= 0then the quotientf /g is continuous atx0.

5. If f : Rm and g : Rm are continuous at x0 then so isthe dot product f g.

6. If f : R3 andg : R3 are continuous atx0 then so is thecross product f g.

7. Iff : R andg : Rm are continuous atx0 then so is thescalar productfg.

Finally, we define a pretty obvious concept that will be important later on.

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(e)

lim(x,y)(0,0)x2y

x4 + y2 .

Problem 9.3. Consider the limit

lim(x,y)(0,0)

x4y4

(x4 + y2)3 .

What happens as (x.y) (0, 0) along the line y = x? What happens alongthe curvey = x2? What can we say about the general limit?

Problem 9.4. For aR define

f(x, y) = sin(x2+y2)

x2+y2 , (x, y)= (0, 0),

a, (x, y) = (0, 0).Is there any choice ofa for which the function is continuous on the whole plane?

Problem 9.5. For aR define

f(x, y) =

e

1(xy)2 x=y,

a x= y

Is there any choice ofa for which the function is continuous on the whole plane?

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Chapter 10

Functions from R to Rn

In the next three chapters, we are going to examine functions involving multiplevariables.

In Chapter 10 (the current chapter) we examine functions whose domainis R and whose range is Rn.

In Chapter 11 we examine functions whose domain is Rn and whose rangeis R.

In Chapter 12 we examine functions whose domain is Rn and whose rangeis Rm.

In these precalculus chapters we will be mostly interested in terminology andvisualization of these functions.

Readers have probably encountered functions from R to Rn before, at leastwhen studying ordinary differential equations. It is typical to think of theindependent variable (in the domain) as time, while the dependent variable canbe the position in space or other quantities (such as temperature, pressure andvolume of a quantity of gas in a cylinder). As we shall see in the remainderof the book, because the domain is simple and one-dimensional, the calculus ofsuch functions is simple as well1.

If we think of the domain of this type of function as time and the range asspace, we think of the motion described by the function sweeping out a curve.The following definitions may seem to draw a lot of overly fine distinctions, butthese distinction can be useful in many situations.

1In fact, the more advanced subject of semigroups considers the calculus of functions

whose domain is R but whose range is infinite-dimensional. While this presents some sig-nificant technical challenges, the bottom line is that the results of calculus and ordinarydifferential can readily be extended to these functions.

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-1-0.5

0

0.5 1

x

-1

-0.5

0

0.5

1y

0

1

2

3

z

1

-0.5

0

0.5y

Figure 10.2: The helix generated by h(t) = (cos t, sin t,t/6), t[0, 6]

Example 10.5. The trajectory

h(t) =

(3t 5, 4t + 7, 2t + 6), t[0, 1]

(4t 6, 11, 7t + 11), t(1, 2]traverses a continuous, piecewise linear curve from initial point (

5, 7, 6) to

the point (2, 11, 4) and then proceeding to the terminal point (2, 11, 3). (SeeFigure 10.3.)

Definition 10.6. We say that a trajectory is a cyclic if its initial andterminal points are the same. A trajectory is simpleif it occupies no pointtwice, except that, perhaps, it could be cyclic.

Example 10.7. The trajectories r1 and r2 given above are both simple andcyclic. The trajectory

r3(t) = (cos t, sin t), t

[0, 4]

would be cyclic, but not simple since it traces the curve twice.

The concepts of curve and trajectory are very similar, but a curve has onlygeometric information. A third concept, path, contains information aboutorder but no information about speed.

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as an oriented simple curve.

Example 10.9. The trajectories

r1(t) = (cos t, sin t), t[0, 2],and

r4(t) = (cos 4t, sin4t), t[0, /2],are equivalent (using the mapping (t) = 4t). The path of these trajectories (asimple cycle) is simply the unit circle, traced counterclockwise, beginning andending with the point (1, 0).

It is often convenient to go backwards along a path.

Definition 10.10. Let r : [t0, t1]

Rn be a trajectory representing the

simple pathP, The reverse ofP is the path equivalent to the trajectoryr : [0, 1] Rn given by

r(s) = r((1 s)t1+ s t0)

fors[0, 1]. We writePfor the reverse ofP.

Problems

Problem 10.1. Define a function describing a trajectory the traverses a linesegment with initial point (1, 3, 4) and terminal point (2, 1, 6).

Problem 10.2. Define a function describing a trajectory the traverses a circleof radius 3 about the origin in the (x, y)-plane. The path should be simple,cyclic, counterclockwise, and the initial point should be (3, 0).Problem 10.3. Define a function describing a trajectory the traverses a circleof radius 1 about the point (2, 4) in the (x, y)-plane. The path should becyclic, clockwise, and traverse the circle exactly three times. The initial pointshould be the point (2, 3).Problem 10.4. Define a function describing a trajectory the traverses a theellipse

x2

9 +

y2

4 = 1.

The path should be simple, cyclic, counterclockwise, and the initial point shouldbe (0, 2).

Problem 10.5. Define a function describing a trajectory the traverses theintersection of the sphere of radius four in R3 and the plane z = 1. The pathshould be simple and cyclic. You may choose the initial point and the orientationthe trajectory, but should describe these clearly.

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graphed the four level curves of the function determined by

xy = 0,xy = 1,

xy = 1,xy = 2,

xy = 2.

These curves are a pair of lines at level c 0 and hyperbolas at the other levels.

-2

-1

0

1

2

x

-2

-1

0

1

2

y

-4

-2

0

2

4

xy

-2

-1

0

1x

-2

-1

0

1y

Figure 11.3: Graph of the functions g(x, y) = xy with the sections in the planesx= 0,

1 and y = 0,

1.

-3 -2 -1 1 2 3x

-3

-2

-1

1

2

3

y

Figure 11.4: Contour plots of the functiong(x, y) = xy at levels c = 0, 1, 2.

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Chapter 12

Functions from Rn to Rm

In this chapter we consider functions where both the domain and range is mul-tidimensional. Of course, for a functionf : Rn Rm we can simply think of itas a collection ofm scalar fields

f1(x1, x2, . . . , xn)f2(x1, x2, . . . , xn)

...fm(x1, x2, . . . , xn)

.

Lets look at some special cases.

12.1 Vector Fields

A vector field is a function where both the domain and range have the samedimension (higher than one).

Example 12.1. There are many important examples from physics of this typeof function.

At every point in space (R3) there exists a measurable electric fieldandmagnetic field, each of them three-dimensional vectors with a magnitudeand direction.

At every point in space there is a gravitational force per unit mass.Again, this is a three-dimensional vector quantity with a magnitude anddirection.

At every point in a moving fluid velocity field describing the speedand direction of motion. It is common to consider both two and three-dimensional flows. The dimension of the velocity vector would correspondto the dimension of the flow.

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Remark 12.2. Visualization of vector fields is difficult. Even in R2, the graphof a vector field would be a subset ofR4 and essentially impossible to visualize

directly. Instead of this, we content ourselves with the method of graphingthe domain and placing a directed line segment representing the output of thefunction at a selection of points in the domain.

The are several computer programs that will help with this. For instance,Mathematica uses the command PlotVectorField and PlotVectorField3Dwhich are in the PlotField3D package.

Example 12.3. The two-dimensional vector field

f(x, y) = (y, x)

represents a counterclockwise flow about the origin. Figure 12.1 displays aMathematica plot of the type described above. When plotting by hand it is often

useful to plot arrows along the axes and a few other lines to begin. For instanceon thex-axis (y = 0), the output of the function has a zerox-component. Thus,it points in a vertical direction up whenx is positive, down when x is negative.Similarly, the output field points to the left on the positive y-axis and to theright on the negative. A similar analysis works for lines like y = x and y =x.

-2 -1 1 2x

-2

-1

1

2

y

Figure 12.1: Plot of the vector field f(x, y) = (y, x).

The field

g(x, y) = x

x2 + y2 , y

x2 + y2represent a flow toward the origin. (See Figure 12.2.) When hand drawing thefield it would be helpful to note that the output of the field at the point ( x, y)is parallel to the vector (x, y). Since the scalar multiple is negative, the outputpoints to the origin.

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-2 -1 1 2x

-2

-1

1

2

y

Figure 12.2: Plot of the vector field g(x, y) = (

x/x2 + y2, y/x2 + y2).Example 12.4. The three-dimensional vector field

v(x, y, z) = (y, x, z)swirls about the z-axis while flowing toward the xy-plane. (See Figure 12.3.)Compare the first two components to the two dimensional field f above. Thethird component points in the negativez direction toward the xy-plane.

-2

0

2x

-2

0

2y

-5

-2.5

0

2.5

5

z

-2

0

2x

-2

0

2y

Figure 12.3: Plot of the swirling 3-D vector field v(x,y,z) = (y, x, z).

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12.2 Parameterized Surfaces

We can use functions from R2 to R3 to describe two-dimensional parameterizedsurfaces in R3.

Example 12.5. We can describe a plane containing the point X0 = (x0, y0, z0)and parallel to the vectors u= (u1, u2, u3) and v= (v1, v2, v3) using the function

p(s, t) = X0+ su + tv=

x0+ su1+ tv1y0+ su2+ tv2

z0+ su3+ tv3

,

Here (s, t)R2.Example 12.6. We can describe a right circular cylinder of radius r centeredon they-axis using the following function h : [0, 2)

R

R3

h(, s) =

r cos s

r sin

,

Herer >0 is fixed. The xand z coordinates sweep out a circle of radiusr aboutthe origin in thexz-plane using the parameter[0, 2). The parametersRtranslates that circle along the y -axis.

Example 12.7. We can describe a sphere of radius > 0 using the functiong: [0, 2) [0, ] R3

g(, ) = cos sin

sin sin cos

,

where >0 is fixed. (, )[0, 2) [0, ] are standard spherical coordinates. is the angle between the vector g and the positive z-axis. is the anglebetween the projection of h onto the xy-plane and the positive x-axis. Notethat by letting range from 0 to and range from 0 to 2, we can representthe entire sphere. If we tried to represent the sphere as a graph we could onlygive half of the sphere with a single function, e.g.

z = f1(x, y) =

2 x2 y2, x2 + y2 2,

andz = f2(x, y) =

2 x2 y2, x2 + y2 2.

Problems

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-1

-0.5

0

0.5

1

x

-1

-0.5

0

0.5

1

y

-1

-0.5

0

0.5

1

z

-1

-0.5

0

0.5x

1

-0.5

0

0.5y

Figure 12.4: Plot of the parameterized spherical surface.

Problem 12.1. Plot the following vector fields.

(a)f(x, y) = (y, x/2).

(b)g(x, y) = (x, y/2).

(c)u(x, y, z) = (0, y, z).

(d)v(x,y,z) = (z, y, x).

Problem 12.2. Represent a right circular cylinder of radius one about they-axis as a parameterized surface.

Problem 12.3. Represent the ellipsoid described by

x2 + 4y2 + 9z2 = 1

as a parameterized surface. (Hint: Modify the a parameterizations of thesphere.)

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(c) S3 ={(x, y) R2 | x2 + (y 1)2 1}, the disk of radius one about thepoint (0, 1).

Problem 13.2. Describe the following sets of points in R3 as sets of cylindricalcoordinates (r,,z).

(a)S1 ={(x, y, z)R3 |

x2 + y2 z1}, a right circular cone with pointat the origin and base on the plane z = 1.

(b) S2 ={(x, y, z) R3 | (x 1)2 +y2 1, 0 < z < 3.}, a right circularcylinder of height three whose base is the disk of radius one about the point(0, 1) in the xy -plane .

(c) S3 ={(x, y, z)R3 |

3

x2 + y2 z

4 xy2}, a volume betweena cone and the sphere about the origin of radius 2.

Problem 13.3. Describe the following sets of points in R3 as sets of sphericalcoordinates (,,).

(a)S1 ={(x, y, z)R3 |

x2 + y2 z1}, a right circular cone with pointat the origin and base on the plane z = 1.

(b) S2 ={(x, y, z) R3 | (x 1)2 +y2 1, 0 < z < 3.}, a right circularcylinder of height three whose base is the disk of radius one about the point(0, 1) in the xy -plane .

(c) S3 ={(x, y, z)R3 |

3

x2 + y2 z

4 xy2}, a volume betweena cone and the sphere about the origin of radius 2.

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Part II

Differential Calculus of

Several Variables

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Chapter 14

Introduction to Differential

Calculus

In this chapter we examine differential calculus of functions of several variables.In order to get our bearings, let us once again consider a syllabus for theanalogous topics in a course in single variable calculus.

1. The derivative of a function of a single variable is defined as the limit ofdifference quotients

f(x) = limyx

f(y) f(x)y x .

2. The difference quotients are interpreted as the slope of secant lines of thegraph off. The derivative is interpreted as the slope of a tangent line.

3. The tangent line is shown to be the line that best approximates theoriginal graph.

4. Various rules for differentiation of functions are developed (e.g. the prod-uct rule, quotient rule, and chain rule).

5. Higher order derivatives are defined. The second derivative is interpretedgeometrically in terms of concavity of the graph.

6. Various applications, including max-min problems (finding the maximumand minimum values of functions) are discussed. The first and secondderivative tests for maxima and minima are developed.

7. In preparation for the definition of the natural logarithm and exponential,theorems about the invertibility of real-valued functions are developed.

Again, this may not be the best order for a course teaching someone calculusfor the first time. But it is a reasonable way to organize the topics covered. Itwill serve us well as an outline for our attack on the same problems with morecomplicated multivariable functions.

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As one might suspect, simply defining a derivative in multiple dimensions isperhaps the hardest part of the program. Once again, we will see that if the

domain of our functions is one-dimensional, calculus proceeds much as before.However, when we move to a multidimensional domain there are many possiblegeneralizations of the derivative. Indeed, we cover several: partial derivatives,the total derivative matrix, the gradient, the divergence, the curl. Each willhave applications in particular situations.

Other parts of the program proceed much as before with some importantcomplications. Scalars are replaced with vectors and matrices; basic algebra isreplaced with linear algebra.

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Theorem 15.8. Let

Rn be the domain of aC1 vector fieldf :

Rn.

Then through everyx0 there exists exactly

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